\newproblem{lay:2_2_11}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.11}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A$ be an invertible $n\times n$ matrix, and let $B$ be an $n\times p$ matrix. Show that the equation $AX=B$ has a unique solution $X=A^{-1}B$.
}{
  % Solution
	Consider the columns of $X$ and $B$:
	\begin{center}
		$X=\begin{pmatrix}\mathbf{x}_1 & \mathbf{x}_2 & ... & \mathbf{x}_p \end{pmatrix}$\\
		$B=\begin{pmatrix}\mathbf{b}_1 & \mathbf{b}_2 & ... & \mathbf{b}_p \end{pmatrix}$\\
	\end{center}
	The matrix equation $AX=B$ is a simultaneous set of equations:
	\begin{center}
		$A\mathbf{x}_1=\mathbf{b}_1$\\
		$A\mathbf{x}_2=\mathbf{b}_2$\\
		...\\
		$A\mathbf{x}_p=\mathbf{b}_p$\\
	\end{center}
	Since $A$ is invertible, each equation has a unique solution given by
	\begin{center}
		$\mathbf{x}_1=A^{-1}\mathbf{b}_1$\\
		$\mathbf{x}_2=A^{-1}\mathbf{b}_2$\\
		...\\
		$\mathbf{x}_p=A^{-1}\mathbf{b}_p$\\
	\end{center}
	Or what is the same
	\begin{center}
		$\begin{array}{rcl}X&=&\begin{pmatrix}A^{-1}\mathbf{b}_1 & A^{-1}\mathbf{b}_2 & ... & A^{-1}\mathbf{b}_p \end{pmatrix}\\
		   &=&A^{-1}\begin{pmatrix}\mathbf{b}_1 & \mathbf{b}_2 & ... & \mathbf{b}_p \end{pmatrix}\\
		   &=&A^{-1}B\\
			\end{array}$
	\end{center}
}
\useproblem{lay:2_2_11}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
